Question

A positronium consists of an electron and a positron revolving about their common centre of mass. Calculate the separation between the electron and positron in their first excited state.

• A. $0.529\mathring{\text{A}}$
• B. $1.05\mathring{\text{A}}$
• C. $2.116\mathring{\text{A}}$
• D. $4.232\mathring{\text{A}}$

Answer 1

The correct option is D $4.232\mathring{\text{A}}$

We use the concept of reduced mass, so we have $\mu \text{(Reduced mass)}=\frac{m}{2}$

Electrostatic force of attraction provides the necessary centripetal force,

$\Rightarrow \frac{k{e}^2}{r^2}=\mu {\omega }^{2}r$

$\Rightarrow \frac{k{e}^2}{r^2}=\frac{m}{2}{\omega }^{2}r=\frac{m{v}^2}{2r}$

$\Rightarrow v=\sqrt{2k{e}^2/mr}........\left(1\right)$

According to Bohr's postulate, Angular momentum is given by,

$\frac{m}{2}\left(v\right)r=\frac{nh}{2\pi }$

$\Rightarrow v=\frac{nh}{\pi mr}........\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$, we get,

$\Rightarrow r=\frac{n^2{h}^2}{2{\pi }^{2}mk{e}^2}=\left(\frac{h^2}{4{\pi }^{2}mk{e}^2}\right)\times 2n^2$

$\Rightarrow r=\left(0.529\mathring{\text{A}}\right)\times 2n^2$

For first excited state, $n=2$

$\therefore r=8\times 0.529\mathring{\text{A}}=4.232\mathring{\text{A}}$

Hence, option $\left(\text{D}\right)$ is correct